CAT 2023 Slot 2 Solved Paper

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Question : 65 of 66

Marks: +1, -0

If

p^2+q^2-29=2 p q-20=52-2 p q

, then the difference between the maximum and minimum possible value of

(p^3-q^3)

243
486
378
189

Solution:

Given that

2 p q-20=52-2 p q⇒4 p q=72 ⇒ p q=18

→(1)

Now,

p^2+q^2-29=2 p q-20⇒p^2+q^2-2 p q=9 ⇒(p-q)^2=9 ⇒ p-q= ± 3

Also,

p^2+q^2-29=2 p q-20⇒p^2+q^2=2 p q+9=2(18)+9=45

Now,

p^3-q^3=(p-q) (p^2+p q+q^2) =(p-q)(45+18)=(p-q)(63)

When

{\p}- {\q}=-3⇒

The value is

63(-3)=-189

and when

{\p}- {\q}=3⇒

The value is

63(3)=189

The difference

⇒ =189-(-189)=378

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